Bravery

ABSTRACT

“BRAVERY” is an engineering calculation method that offers theoretical solutions rather than numerical solutions, and can be used to simulate the problems encountered in the engineering field, and subsequently to be used in engineering design.

BRIEF SUMMARY

“BRAVERY” is an engineering calculation method that offers theoreticalsolutions rather than numerical solutions, and can be used to simulatethe problems encountered in the engineering field, and subsequently tobe used in engineering design.

BACKGROUND

After the Finite Element Method was created, the engineering fieldentered a more mature and steady stage in terms of engineeringcalculation. However, the analysis results of the Finite Element Methodwere often unsatisfactory to engineering designers, and its accuracy hasyet to be improved. “BRAVERY” is an engineering calculation method thatoffers theoretical solutions rather than numerical solutions, and can beused to simulate the problems encountered in the engineering field, andsubsequently to be used in engineering design. The advantage of BRAVERYis that it can solve any engineering problem if engineers can preciselydefine their deformation curve or deformation surface.

The Finite Element Method requires that engineers must have a basicconcept of the engineering problems in order to perform finite elementanalysis. BRAVERY also has the same requirement, or even a higherrequirement regarding this concept, because engineers can only adopt theBRAVERY calculation method after they have reached an understanding ofthe engineering problems they are going to analyze, to a considerabledegree (i.e., having already seen photos, images or data of similarproblems, and can almost define the deformation curve or deformationsurface).

With the Finite Element Method, if engineers have no basic concept ofthe engineering problems, the error in the engineering analysis resultsmay go unseen, thus leading to the difficulty in determining whether theproblem lies in errors with the engineering model, improper setting ofboundary conditions, element selection error, poor mesh generation, oreven a problem with the finite element software itself. Although BRAVERYhas a higher requirement for engineers, its aim is the same as that ofthe Finite Element Method, namely, to avoid errors in the results of theengineering analysis. Simply stated, the Finite Element Method involvescalculating results in a backwards direction, while BRAVERY involvespredicting results in a forward direction.

DETAILED DESCRIPTION

BRAVERY is an engineering calculation method which creates an energyequivalent formula based on the given deformation curve or deformationsurface to deduce the initial hypothetical variables.

1. BRAVERY is an engineering calculation method which first calculatesthe axial strain and shear strain in various directions with the givendeformation curve or deformation surface equation, and then establishesan equation based on the kinetic energy (K=½ mν²) or the work (W=Fd) ofthe load with the object's strain energy (U=½∫_(V) Eε²dV) to obtain atheoretical energy equivalent formula. Then, numerical methods are usedto solve the equation in order to obtain the desired dimension variablesvia repeated calculations in a process based on trial and error.Finally, based on the material stress as the basis of equation control,the value of the dimension variables is obtained. The following is anexample of a calculation of a high-speed ball hitting a circular plate,with simplified theoretical derivation. Here, the impact kinetic energycan be expressed as:K=½mξ² Suppose that the supporting object behind the circular plate is aflexible entity; when the ball hits the circular plate from the front,the plastic deformation curve of the latter accords with the Gaussiandistribution curve: μ=0. Before proportional magnification, its plasticdeformation curve equation can be expressed as:${\delta_{A}(x)} = \frac{^{- \frac{x^{2}}{2\sigma^{2}}}}{\sigma \sqrt{2\pi}}$After the circular plate deforms plastically, suppose its maximumallowable midpoint deformation is δ; the proportional magnification isexpressed as:$X = {\frac{\delta}{\left. \frac{^{- \frac{x^{2}}{2\sigma^{2}}}}{\sigma \sqrt{2\pi}} \right|_{x = 0}} = {{\delta\sigma}\sqrt{2\pi}}}$Based on the above magnification, the theoretical plastic deformationcurve equation of the circular plate can be expressed as:${\delta_{T}(x)} = {{X\frac{^{- \frac{x^{2}}{2\sigma^{2}}}}{\sigma \sqrt{2\pi}}} = {\frac{{\delta\sigma}\sqrt{2\pi}^{- \frac{x^{2}}{2\sigma^{2}}}}{\sigma \sqrt{2\pi}} = {\delta }^{- \frac{x^{2}}{2\sigma^{2}}}}}$The second derivative is performed on dx to acquire the curvature κ ofthe plastic deformation curve, and then to obtain its radius ofcurvature through reciprocity. The radius of curvature of the plasticdeformation curve can be expressed as:$\rho = {\frac{1}{\kappa} = {\frac{1}{\frac{^{2}}{x^{2}}\left( {\delta }^{- \frac{x^{2}}{2\sigma^{2}}} \right)} = \frac{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}}{\delta \left( {x^{2} - \sigma^{2}} \right)}}}$Without considering the positive compression of the circular plate, theoverall strain energy is selected in a relatively conservative way; thusthe effect of compression strain ε_(γ) on the strain energy of thecircular plate is ignored. Suppose the distance between the neutral axisof the circular plate and its periphery is γ; based on the above-definedradius of curvature ρ, and since the strain in the two-way symmetricdirections is the same, the longitudinal and lateral strain ε_(x), ε_(z)can be defined as:$\varepsilon_{x} = {\varepsilon_{z} = {{\frac{y}{\rho}\left( {1 - v} \right)} = {\frac{y\left( {1 - v} \right)}{\frac{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}}{\delta \left( {x^{2} - \sigma^{2}} \right)}} = \frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}}}}}$When the ball penetrates the circular plate, the shear strainconcentrates on a tiny area on the projection plane of the ball, makinglittle contribution to the strain energy in other areas; the strainenergy contribution of the shear strain on the projection plane of theball can therefore be ignored. Hard titanium compound with evaporationcoating may be used on the upper and lower surface of the substrate toresist the impact of the ball so that the circular plate can deformplastically with better effect. Based on the two-way strain of thecircular plate defined in the above equation, its strain energy can beexpressed as: $\begin{matrix}{U = {{\frac{1}{2}{\int_{V}{E\; \varepsilon^{2}{V}}}} = {{\frac{1}{2}{\int_{V_{i}}{E_{i}\varepsilon^{2}{V_{i}}}}} +}}} \\{{{\frac{1}{2}{\int_{V_{o}}{E_{0}\varepsilon^{2}{V_{o}}}}} + {\frac{1}{2}{\int_{V_{o}}{E_{o}\varepsilon^{2}{V_{o}}}}}}} \\{= {{\frac{E_{i}}{2}{\int_{V_{i}}{\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}^{T}\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}{V_{i}}}}} +}} \\{{{\int_{V_{o}}{\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}^{T}\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}{V_{o}}}} +}} \\{{\frac{E_{o}}{2}{\int_{V_{o}}{\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}^{T}\begin{Bmatrix}\varepsilon_{x} \\\varepsilon_{z}\end{Bmatrix}{V_{o}}}}}} \\{= {E_{i}{\int_{- \frac{t}{2}}^{\frac{t}{2}}{\int_{0}^{R}{2\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {\delta \; ^{- \frac{x^{2}}{2\sigma^{2}}}} \right)} \right\rbrack^{2}}}}}}} \\{{{\left( \frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}{y}} +}} \\{{E_{o}{\int_{\frac{t}{2}}^{\frac{t}{2} + c}{\int_{0}^{R}{2\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {\delta \; ^{- \frac{x^{2}}{2\sigma^{2}}}} \right)} \right\rbrack^{2}}}}}}} \\{{{\left( \frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}{y}} +}} \\{{E_{o}{\int_{{- \frac{t}{2}} - c}^{- \frac{t}{2}}{\int_{0}^{R}{2\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {\delta \; ^{- \frac{x^{2}}{2\sigma^{2}}}} \right)} \right\rbrack^{2}}}}}}} \\{{\left( \frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}{y}}} \\{= \frac{{t^{3}E_{i}} + {2{c\left( {{4c^{2}} + {6{ct}} + {3t^{2}}} \right)}E_{o}}}{6}} \\{{\int_{0}^{R}{\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {\delta \; ^{- \frac{x^{2}}{2\sigma^{2}}}} \right)} \right\rbrack^{2}}}}} \\{{\left( \frac{\; {{\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}}}\end{matrix}$ According to the strain energy of the circular plate andthe impact kinetic energy of the ball, an equivalent formula can beestablished as:${\frac{1}{2}{mv}^{2}} = {\frac{{t^{3}E_{i}} + {2{c\left( {{4c^{2}} + {6{ct}} + {3t^{2}}} \right)}E_{o}}}{6}{\int_{0}^{R}{\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {\delta }^{- \frac{x^{2}}{2\sigma^{2}}} \right)} \right\rbrack^{2}}\left( \frac{\; {{\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}}}}$Through reorganization, the formula can be re-expressed as:${{t^{3}E_{i}} + {2{c\left( {{4c^{2}} + {6{ct}} + {3t^{2}}} \right)}E_{o}}} = \frac{3{mv}^{2}}{\int_{0}^{R}{\pi \; x\sqrt{1 + {\left\lbrack {\frac{\;}{x}\left( {\delta }^{- \frac{x^{2}}{2\sigma^{2}}} \right)} \right\rbrack^{2}\left( \frac{\; {{\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \right)^{2}{x}}}}}$The above is a theoretical calculation formula for a ball hitting acircular plate, which adopts calculation based on trial and error. Thefollowing is an example of how to solve the above equation. To avoidnumerical errors, N and mm are taken. The plastic deformation curve ofthe circular plate can be simulated with the Gaussian distributioncurve: σ=6. The measured impact energy of AK47, M16 and .45 pistol is2460J, 1796J and 835J, which are taken as the target to be achieved inthe circular plate design. The theoretical calculation formulae relatedto the terms and orders of kinetic energy are then: 3 m_(c)ν_(c)²=14760J, 10776J, 5010J=14760000, 10776000, 5010000 N mm. If a medal iscapable of stopping a bullet, it will greatly enhance the bravery ofthose wearing it. The following therefore offers a design for a medalcapable of stopping said bullet. Under given conditions, the diameter ofthe medal is set as 44 mm (R=22 mm), the substrate is Titanium alloyTi6Al4V used for casting (with elastic modulus E_(i)=113.8 GPa,Poisson's ratio ν=0.342, and yield strength σ_(y)=880 MPα), the upperand lower surface are evaporation coated (c=8 μm) with titanium carbideTiC (with elastic modulus E_(o)=448 GPa, and Mohs hardness Mohs=9.5). Todetermine the thickness of the medal, it is necessary to repeatedly makerectifications, and verify whether the peripheral stress value withinthe width of the bullet has reached the yield strength, in order toensure that the bullet will not penetrate through the medal. Throughrepeated trial and error, the following three sets of calculating datawere obtained:A.  AK 47  (w = 7.62  mm, δ = 1.735  mm, t = 13.079  mm)${\int_{0}^{22}{\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {1.735\mspace{11mu} ^{- \frac{x^{2}}{2 \times 6^{2}}}} \right)} \right\rbrack^{2}}\left( \frac{1.735\left( {x^{2} - 6^{2}} \right)\left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{x^{2}}{2 \times 6^{2}}}} \right)^{2}{x}}} = 0.057149024133450785$${{t^{3} \times 113800} + {2 \times 0.008 \times \left( {{4 \times 0.008^{2}} + {6 \times 0.008 \times t} + {3 \times t^{2}}} \right) \times 448000}} = \frac{14760000}{0.057149024133450785}$t = 13.078741511388055  mm$\sigma_{i,\max} = {{\frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \cdot E_{i}} = {{\frac{\frac{13.078741511388055}{2} \times 1.735 \times \left\lbrack {\left( \frac{7.62}{2} \right)^{2} - 6^{2}} \right\rbrack \left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{{(\frac{7.62}{2})}^{2}}{2 \times 6^{2}}}} \times 113800} = {{{- 880.2105377936211} \cong \sigma_{y}} = {880\mspace{14mu} {MPa}}}}}$B.  M 16  (w = 5.56  mm, δ = 1.200  mm, t = 15.081  mm)${\int_{0}^{22}{\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {1.200\mspace{11mu} ^{- \frac{x^{2}}{2 \times 6^{2}}}}\; \right)} \right\rbrack^{2}}\left( \frac{1.200\left( {x^{2} - 6^{2}} \right)\left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{x^{2}}{2 \times 6^{2}}}} \right)^{2}{x}}} = 0.027264640739074912$${{t^{3} \times 113800} + {2 \times 0.008 \times \left( {{4 \times 0.008^{2}} + {6 \times 0.008 \times t} + {3 \times t^{2}}} \right) \times 448000}} = \frac{10776000}{0.027264640739074912}$t = 15.081134481897694  mm $\sigma_{i,\max} = {{\frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \cdot E_{i}} = {{\frac{\frac{15.081134481897694}{2} \times 1.200 \times \left\lbrack {\left( \frac{5.56}{2} \right)^{2} - 6^{2}} \right\rbrack \left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{{(\frac{5.56}{2})}^{2}}{2 \times 6^{2}}}} \times 113800} = {{{- 880.335276252948}\mspace{11mu} \cong \sigma_{y}} = {880\mspace{14mu} {MPa}}}}}$C.  .45  pistol  (w = 11.5  mm, δ = 16.413  mm, t = 1.788  mm)${\int_{0}^{22}{\pi \; x\sqrt{1 + \left\lbrack {\frac{\;}{x}\left( {16.413\mspace{11mu} ^{- \frac{x^{2}}{2 \times 6^{2}}}}\; \right)} \right\rbrack^{2}}\left( \frac{16.413\left( {x^{2} - 6^{2}} \right)\left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{x^{2}}{2 \times 6^{2}}}} \right)^{2}{x}}} = 6.96193561779727$${{t^{3} \times 113800} + {2 \times 0.008 \times \left( {{4 \times 0.008^{2}} + {6 \times 0.008 \times t} + {3 \times t^{3}}} \right) \times 448000}} = \frac{5010000}{6.96193561779727}$t = 1.7878006817497325  mm$\sigma_{i,\max} = {{\frac{y\; {\delta \left( {x^{2} - \sigma^{2}} \right)}\left( {1 - v} \right)}{\sigma^{4}^{\frac{x^{2}}{2\sigma^{2}}}} \cdot E_{i}} = {{\frac{\frac{1.7878006817497325}{2} \times 16.413 \times \left\lbrack {\left( \frac{11.5}{2} \right)^{2} - 6^{2}} \right\rbrack \left( {1 - 0.342} \right)}{6^{4} \times ^{\frac{{(\frac{11.5}{2})}^{2}}{2 \times 6^{2}}}} \times 113800} = {{{- 879.9750808686258}\mspace{11mu} \cong \sigma_{y}} = {880\mspace{14mu} {MPa}}}}}$By solving the above equations, the obtained thicknesses of medalcorresponding to AK47, M16 and .45 pistol are: t=13.079 mm, 15.081 mm,and 1.788 mm, respectively. Through calculations, it is known that thebullet of an AK47 possesses more energy than that of an M16, but alsothat the bullet is wider than that of the latter. That is why it isharder for an AK47, but also that the bullet is wider than that of thelatter. That is why it is harder for an AK47 to penetrate through themedal. A .45 pistol is mainly used for close range shooting, and itsbullet has low energy and large width; this is the main reason that itspenetration power is far lower than that of either an AK47 or an M16.With a reasonable thickness of medal, a .45 pistol may be taken as thedesign target to realize the increase in bravery. To ensure that themedal can effectively withstand the impact of a bullet, a shootingsimulation is carried out with real guns and bullets. Then by correctingthe gap between theory and actual measurement using statisticalregression, a more complete design of bullet-stopping medal can beobtained. Sign List K Ball's impact kinetic energy (N mm) m Ball's mass(g) ν Ball's velocity (m/s) W Work (Nmm) F Force (N) d Displacement (mm)w Ball's width (mm) U Strain energy of the circular plate (N mm)δ_(A)(x) Plastic deformation curve equation of the circular plate (mm)δ_(T)(x) Theoretical plastic deformation curve equation of the circularplate (mm) X Proportion magnification of the plastic deformation; nounits σ Parameter of the plastic deformation curve (mm) δ Maximummidpoint plastic deformation (mm) κ Curvature of the plastic deformationcurve (mm⁻) γ Distance between the neutral axis and the periphery (mm) ρCurvature radius of the plastic deformation curve (mm) ε_(x),E ε_(z)Longitudinal and lateral strains of the circular plate; no units νPoisson's ratio of the substrate material; no units E_(i) Elasticmodulus of the substrate material (N/mm²) E_(o) Elastic modulus of theevaporation coated material (N/mm²) R Design radius of the circularplate (mm) t Substrate thickness of the circular plate (mm) cEvaporation coated thickness of the circular plate (mm)
 2. When BRAVERYis applied to materials which have already been deformed plastically(for example, pressing processing), it is necessary to evaluate itsamount of plastic deformation and the residual stress in order to reducethe allowable stress value used in calculation. Otherwise, pressing maycause the material to undergo inner plastic deformation, leading to anover-estimation of the design strength, and resulting in design failure.3. BRAVERY may be developed into engineering software for calculationpurposes, to shorten the calculation time spent by engineers. Theengineering software should provide, and allow manual creation of,commonly used deformation curve and deformation surface modules forengineers to quickly establish theoretical energy equivalent formulaefor the engineering problems encountered, and then solve by repeatedtrial and error process using the built-in numerical methods to derivethe value of the initial hypothetical variables. When the results areobtained, the strain and stress values of each block are calculated inaccordance with the strain equation, to generate a graph for engineersto examine and verify. The definitions of the deformation curve anddeformation surface are derived from the curve or curved surface, andthrough the equation definition for producing lines through toolfunctions, the deformation curve and deformation surface equation can beprecisely defined.